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June 5th, 2014, 03:48 AM   #1
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log

log15/log6=a

log18/log12=b

find log25/log24
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June 5th, 2014, 05:54 AM   #2
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$\displaystyle a = \frac{\log 15}{\log6} = \frac{\log 3 + \log 5}{\log 3 + \log 2}$
$\displaystyle b = \frac{\log 18}{\log 12} = \frac{\log 2 + 2\log 3}{\log 3 + 2\log 2}$

Rearrange a to make $\displaystyle \log 5$ the subject:

$\displaystyle \log 5 = a(\log 3 + \log 2) - \log 3$

Rearrange b to make $\displaystyle \log 3$ the subject:

$\displaystyle \log 3 = \frac{2b-1}{2-b} \cdot \log 2$

$\displaystyle \frac{\log 25}{\log 24}= \frac{2\log 5}{3\log 2 + \log 3}$
$\displaystyle = \frac{2a(\log 3 + \log 2) - 2\log 3}{3\log 2 + \log 3}$
$\displaystyle = \frac{2a\left(\frac{2b-1}{2-b}\log 2 + \log 2\right) - 2\left(\frac{2b-1}{2-b}\log 2\right)}{3\log 2 + \left(\frac{2b-1}{2-b}\log 2\right)}$
$\displaystyle = \frac{2a\left(\frac{2b-1}{2-b}\log 2\right) + 2a\log 2 - 2\left(\frac{2b-1}{2-b}\log 2\right)}{3\log 2 + \left(\frac{2b-1}{2-b}\log 2\right)}$
$\displaystyle = \frac{2a\left(\frac{2b-1}{2-b}\right) + 2a - 2\left(\frac{2b-1}{2-b}\right)}{3 + \left(\frac{2b-1}{2-b}\right)}$
$\displaystyle = \frac{2a(2b-1) + 2a(2-b) - 2(2b-1)}{3(2-b) + 2b-1}$
$\displaystyle = \frac{4ab -2a+4a-2ab-4b+2}{6-3b+2b-1}$
$\displaystyle = \frac{2ab+2a-4b+2}{5-b}$
$\displaystyle \frac{2(ab+a-2b+1)}{5-b}$
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